Answer: \(R=(35,-18)\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

When \(x=5\),

\(y=(5(5)+2)^{2/3}=27^{2/3}=9.\)

So \(P=(5,9)\).

Differentiate:

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac23(5x+2)^{-1/3}\cdot5=\frac{10}{3}(5x+2)^{-1/3}.\)

At \(x=5\),

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{10}{3}\cdot\frac13=\frac{10}{9}.\)

So the normal has gradient \(-\frac{9}{10}\). Its equation is

\(y-9=-\frac{9}{10}(x-5).\)

The point \(Q\) also lies on \(x+y=11\), so \(y=11-x\). Substitute into the normal equation:

\(11-x-9=-\frac{9}{10}(x-5).\)

Thus

\(2-x=-\frac{9}{10}x+\frac{9}{2}.\)

Multiplying by \(10\),

\(20-10x=-9x+45,\)

so \(x=-25\). Hence \(y=36\), and \(Q=(-25,36)\).

Since \(Q\) lies on the normal and the tangent is perpendicular to the normal at \(P\), the tangent is the perpendicular bisector of \(QR\). Therefore \(P\) is the midpoint of \(QR\).

So

\(R=2P-Q=(10,18)-(-25,36)=(35,-18).\)

This gives the required answer: \(R=(35,-18)\).